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19. Merge Intervals

mediumAsked at Coinbase

Collapse overlapping trading windows into consolidated ranges — Coinbase uses interval merging to evaluate how engineers reason about time-series events like order-book snapshots and candlestick aggregation.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals and return an array of the non-overlapping intervals that cover all input intervals.

Constraints

  • 1 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= start_i <= end_i <= 10^4

Examples

Example 1

Input
intervals = [[1,3],[2,6],[8,10],[15,18]]
Output
[[1,6],[8,10],[15,18]]

Explanation: [1,3] and [2,6] overlap, merge to [1,6].

Example 2

Input
intervals = [[1,4],[4,5]]
Output
[[1,5]]

Explanation: Touching endpoints count as overlapping.

Approaches

1. Brute force (pairwise check)

Repeatedly scan all pairs and merge any that overlap until no more merges occur. O(n^2) per pass.

Time
O(n^2)
Space
O(n)
function mergeIntervals(intervals) {
  let changed = true;
  while (changed) {
    changed = false;
    const next = [];
    const used = new Array(intervals.length).fill(false);
    for (let i = 0; i < intervals.length; i++) {
      if (used[i]) continue;
      let [s, e] = intervals[i];
      for (let j = i + 1; j < intervals.length; j++) {
        if (used[j]) continue;
        const [s2, e2] = intervals[j];
        if (s2 <= e && e2 >= s) {
          s = Math.min(s, s2);
          e = Math.max(e, e2);
          used[j] = true;
          changed = true;
        }
      }
      next.push([s, e]);
    }
    intervals = next;
  }
  return intervals;
}

Tradeoff:

2. Sort then linear scan (optimal)

Sort by start time; iterate once, extending the current interval or pushing a new one when no overlap.

Time
O(n log n)
Space
O(n)
function merge(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  const result = [intervals[0]];
  for (let i = 1; i < intervals.length; i++) {
    const last = result[result.length - 1];
    if (intervals[i][0] <= last[1]) {
      last[1] = Math.max(last[1], intervals[i][1]);
    } else {
      result.push(intervals[i]);
    }
  }
  return result;
}

Tradeoff:

Coinbase-specific tips

Coinbase evaluates whether you immediately sort before scanning — the interview equivalent of normalising a time-series before aggregating OHLC candles. They also watch for off-by-one errors on touching endpoints (start_i == end_j merges), which maps directly to fence-post bugs in settlement-window calculations.

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